Buffering capacity experiment

We are doing a experiement where we are testing the buffering capacity of soils. The text says to use Ca(OH)2, but we only have NaOH, which i plan to substitute. I just wanted to check that the solution i make up will have the equivalent OH.

2. Relevant equations
making up 0.01M solution of Ca(OH)2 which is 0.07g per litre

By inspection your answer is way off. You are saying that of the 0.07 grams of Ca(OH)2 present in a liter that 0.34 grams of that is due to the counterion, OH? How can you have 0.34 grams out of only 0.07 grams?

Ahh, great! You want 0.01 moles per liter of Ca(OH)2 which is 0.02 equivalents of OH- per liter. To get that from NaOH you need (0.02moles/L OH-) X (40 g NaOH/mole) which is 0.8 g/L NaOH.... same answer as yours.